To determine the time it takes for the activity of a sample of Radium-223 (with a half-life of 11.4 days) to decrease to 2.00% of its initial value, follow these steps:
1. Understand the concept of half-life: The half-life of a radioactive substance is the time it takes for half of the substance to decay. For Radium-223, the half-life is 11.4 days.
2. Set up the decay formula: The decay of a radioactive substance over time can be described using the formula:
[tex]\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{\text{half}}}} \][/tex]
where:
- [tex]\( N(t) \)[/tex] is the remaining quantity of the substance at time [tex]\( t \)[/tex],
- [tex]\( N_0 \)[/tex] is the initial quantity of the substance,
- [tex]\( t_{\text{half}} \)[/tex] is the half-life,
- [tex]\( t \)[/tex] is the time elapsed.
3. Determine the problem's parameters:
- [tex]\( t_{\text{half}} = 11.4 \)[/tex] days
- The activity decreases to 2.00% of its initial value, meaning [tex]\( \frac{N(t)}{N_0} = 0.02 \)[/tex].
4. Express the decay in logarithmic form:
[tex]\( N(t) / N_0 = 0.02 \)[/tex]
[tex]\[ \left( \frac{1}{2} \right)^{\frac{t}{11.4}} = 0.02 \][/tex]
5. Isolate [tex]\( t \)[/tex] by applying the natural logarithm (ln) to both sides of the equation:
[tex]\[ \ln \left( \left( \frac{1}{2} \right)^{\frac{t}{11.4}} \right) = \ln(0.02) \][/tex]
Using the logarithmic identity [tex]\( \ln(a^b) = b \ln(a) \)[/tex]:
[tex]\[ \frac{t}{11.4} \cdot \ln\left( \frac{1}{2} \right) = \ln(0.02) \][/tex]
6. Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{11.4 \cdot \ln(0.02)}{\ln(1/2)} \][/tex]
7. Calculate the value:
Using the known values of natural logarithms:
- [tex]\( \ln(0.02) \approx -3.912 \)[/tex]
- [tex]\( \ln(1/2) \approx -0.693 \)[/tex]
[tex]\[ t \approx \frac{11.4 \cdot (-3.912)}{(-0.693)} \][/tex]
[tex]\[ t \approx \frac{-44.5768}{-0.693} \][/tex]
[tex]\[ t \approx 64.34 \][/tex]
Thus, the time it takes for the activity of a sample of Radium-223 to decrease to 2.00% of its initial value is approximately 64.34 days.
